Given that,
D1 = 60cm
D2 = 24cm
N2 = 300rpm µ = 0.3
X = 3m = 300cm
P = 3.75KW
Safe Tension = Maximum tension = 100N/cm width = 100b N b = width of belt
Tmax = 100b ...(i)
Let θ = Angle of contact
Sinα = (r1 – r2)/X = (30 – 12)/300 ; α = 3.45°, ...(ii)
= Π – 2α = (180 – 2 X 3.45) = 173.1°
= (173.1°)X Π/180 = 3.02 rad ......(iii)
Now,
Using the relation, Ratio of belt tension = T1/T2 = eµθ = e(0.3)(3.02)
T1 = 2.474 T2 ......(iv)
Now,
V = πDN/60 m/sec, D is in meter and N is in RPM
= 3.14 X (0.24)(300)/60 = 3.77m/sec ......(v)
Power Transmitted (P) = (T1 – T2).v/1000 Kw
3.75 = (T1 – T2)X 3.77/1000
T1 – T2 = 994.7N .......(vi)
From relation (iv) and (v), we get:
T1 = 1669.5N ...........(vii)
T2 = 674.8N ...........(viii)
(i) For width of the belt
But T1 = Tmax = 100b; 1669.5 = 100b; b = 16.7cm
(ii) For initial tension in the belt
Let To = initial tension in the belt
To = (T1 + T2)/2
= (1669.5 + 674.8)/2
To = 1172.15N
(iii) For length of belt
Putting all the value, we get
L = 7.33m.